Fluid Mechanics, Particle Size, Bouyouncy Force

[Aqib409]

1. What should be the size of object/Particle (sphere) so that it should remain suspended (neither it should sink nor it should float)? container dimensions 30*25*25, density of fluid 0.9g/cm3 , density of particle 0.86g/cm3 , particle is coated with aluminium having density of 2.07g/cm3.(if radius is assumed then find out at what height it remains suspended in a container)2. area of sphere, area of hollow sphere , bouyouncy force, conditions for floating,sinking and suspension.3. Fb = Fg for suspended particle , where Fb = density of liquid * volume of displace water * gravitaional acceleration
Fg = mass of particle * gravitational acceleration

For finding Height we can use P = W * A * H
where
p = pressure acting on a submerged body
A = surface area of particle
H = height of suspended particle

And i m feeling difficulty in finding Pressure to determine height .

 

[haruspex]

1. What should be the size of object/Particle (sphere) so that it should remain suspended (neither it should sink nor it should float)? container dimensions 30*25*25, density of fluid 0.9g/cm3 , density of particle 0.86g/cm3 , particle is coated with aluminium having density of 2.07g/cm3.(if radius is assumed then find out at what height it remains suspended in a container)2. area of sphere, area of hollow sphere , bouyouncy force, conditions for floating,sinking and suspension.3. Fb = Fg for suspended particle , where Fb = density of liquid * volume of displace water * gravitaional acceleration
Fg = mass of particle * gravitational acceleration

For finding Height we can use P = W * A * H
where
p = pressure acting on a submerged body
A = surface area of particle
H = height of suspended particle

And i m feeling difficulty in finding Pressure to determine height .

Use Archimedes' Principle.
However, there are a couple of strange things here. I would have expected one radius to be given (the uncoated particle, the coated particle, or the coating thickness) and the task wouid be to find the other two.
The other puzzle is the question about height of flotation. For that to make sense we wouid have to treat either the fluid or the particle as compressible, but we have no information on the bulk modulus of either.

 

[Aqib409]

Use Archimedes' Principle.
However, there are a couple of strange things here. I would have expected one radius to be given (the uncoated particle, the coated particle, or the coating thickness) and the task wouid be to find the other two.
The other puzzle is the question about height of flotation. For that to make sense we wouid have to treat either the fluid or the particle as compressible, but we have no information on the bulk modulus of either.

yes i am using archimedes principle , i have assumed radius as well e.g if we assume that we have wood sphere coated with aluminium and radius of wood particle is 1.9cm and aluminium coating will be taken as hollow sphere having internal radius of 1.9cm and outer radius is 2cm (mean coating thickness = 0.1cm) then at this point Fb = Fg which means particle will suspend , as we have assumed radius values now we have to find height at which it will be suspended in container for this purpose i am using this formula
p = specific weight of liquid * surface Area of particle * height
for this i have value of specific weight of liquid and surface area of particle but height H and pressure acting on particle when it is in suspended state are unknown
so the issue i am facing now is how should i find that pressure to find height?

 

[haruspex]

yes i am using archimedes principle

It does not seem to me that you are. What does the principle state? How have you applied it?

p = specific weight of liquid * surface Area of particle * height

Where did you get this from? What do you think the resulting p represents?
I can't think of any use for such a formula.

 

[Aqib409]

It does not seem to me that you are. What does the principle state? How have you applied it?

Where did you get this from? What do you think the resulting p represents?
I can't think of any use for such a formula.

Fluid Mechanics by RS Khurmi
Chapter # 3 : Hydrostatics
Where he used this formula to get pressure acting on a immersed surface .

P is pressure acting on a immersed surface.

 

[Aqib409]

It does not seem to me that you are. What does the principle state? How have you applied it?

I can't think of any use for such a formula.

then which formula should i use to find the height of suspended particle in a container ?

 

[haruspex]

Fluid Mechanics by RS Khurmi
Chapter # 3 : Hydrostatics
Where he used this formula to get pressure acting on a immersed surface .

P is pressure acting on a immersed surface.

The P=WAH equation doesn't make any sense to me. What is W?
If an incompressible liquid has density ρ then at depth h the pressure is ρgh greater than at the surface. The force that exerts on a horizontal area A is ρghA. For a surface that is not all at the same depth the pressure varies across it.
But none of this is needed if you use Archimedes' Principle.

Please state Archimedes' Principle and explain in what way you believe you have used it.

then which formula should i use to find the height of suspended particle in a container ?

As I wrote, if neither the fluid nor the body are compressible there is no way to find at what height a fully submersed body will float. If the body floats but is fully submerged it is said to have neutral buoyancy, and in theory will happily float at any depth.
In practice, everything is at least a little compressible. If the body is the more compressible then it will only float neutrally at one depth, but this is unstable. If it rises just slightly it will continue up, while if it sinks at all it will continue down.
If the fluid is more compressible than the body then there will be a depth at which it is stable.

But let's put all that aside for now and concentrate on the first part. Suppose the inner sphere has density ρ₁, radius r₁, and the coating has density ρ₂, thickness r₂. If that floats neutrally in a fluid of density ρ_f, what equation can you write using Archimedes' Principle?

 

[Aqib409]

The P=WAH equation doesn't make any sense to me. What is W?
If an incompressible liquid has density ρ then at depth h the pressure is ρgh greater than at the surface. The force that exerts on a horizontal area A is ρghA. For a surface that is not all at the same depth the pressure varies across it.
But none of this is needed if you use Archimedes' Principle.

Please state Archimedes' Principle and explain in what way you believe you have used it.

As I wrote, if neither the fluid nor the body are compressible there is no way to find at what height a fully submersed body will float. If the body floats but is fully submerged it is said to have neutral buoyancy, and in theory will happily float at any depth.
In practice, everything is at least a little compressible. If the body is the more compressible then it will only float neutrally at one depth, but this is unstable. If it rises just slightly it will continue up, while if it sinks at all it will continue down.
If the fluid is more compressible than the body then there will be a depth at which it is stable.

But let's put all that aside for now and concentrate on the first part. Suppose the inner sphere has density ρ₁, radius r₁, and the coating has density ρ₂, thickness r₂. If that floats neutrally in a fluid of density ρ_f, what equation can you write using Archimedes' Principle?

"w" is the specific weight of fluid
Where as i have mentioned above i.e at assumed radius Fb = Fg
Where
Fb = density of fluid * volume of fluid displaced * gravitational acceleration
Fg = mass of body * gravitational acceleration

When we attain Fb = Fg body will suspend at any height in fluid which is archimedes principle... Till this point we are right because we have attained suspension but now how we will find out the height?

 

[haruspex]

When we attain Fb = Fg body will suspend at any height in fluid which is archimedes principle

Ok, but have you answered the question about the size of the particle?

how we will find out the height?

As I explained, and as you yourself have written, it will suspend at any height. There is no information by which to find a height.

I cannot understand why the question says "if a radius is assumed". Is this a translation?

 

[Aqib409]

Ok, but have you answered the question about the size of the particle?

As I explained, and as you yourself have written, it will suspend at any height. There is no information by which to find a height.

I cannot understand why the question says "if a radius is assumed". Is this a translation?

Yes sir, i have answered the questions if i assume
inner radius = 0.024m
Outer radius = 0.025m
Then
Fb = Fg

But now we have to find the suspension height of the particle (of assumed radius)

Or let's make it simple i need to find the pressure acting on a suspended particle if these parameters are known

Density of fluid = 900kg/m3
Density of coating material = 2700 kg/m3
Density of inner sphere material = 630kg/m3
Container dimensions = 30 * 25 * 25
Radius of coating (hollow shell,outer) = 0.025m
Radius of inner sphere = 0.024m

 

[haruspex]

But now we have to find the suspension height of the particle (of assumed radius)

As I keep posting, there is no way to do that. If there were any particular height then the buoyancy force would only match the weight at that height, and you know that is not so.

Or let's make it simple i need to find the pressure acting on a suspended particle if these parameters are known

First, that has nothing to do with finding the suspension height.
Secondly, the pressure is different in magnitude and direction on different parts of the surface. You could calculate exactly how it varies over the surface, and from that you could deduce the net force from these pressures, but that would just be a complicated way of arriving at what Archimedes has already told you.

A question... how did you arrive at those two radii? You chose one of them arbitrarily, right?

 

[Aqib409]

As I keep posting, there is no way to do that. If there were any particular height then the buoyancy force would only match the weight at that height, and you know that is not so.

First, that has nothing to do with finding the suspension height.
Secondly, the pressure is different in magnitude and direction on different parts of the surface. You could calculate exactly how it varies over the surface, and from that you could deduce the net force from these pressures, but that would just be a complicated way of arriving at what Archimedes has already told you.

A question... how did you arrive at those two radii? You chose one of them arbitrarily, right?

Ok what first i did is i have assumed that i have a solid ball having some radius, with the help of that radius i calculated volume , then with density and volume i found mass of that ball

Then i used
Fb = Fg

Till this point the values of these two doesn't comes equal to each other, so now i have to etract some mass from that solid ball and mass should be equal to that with the help of which we attain equality between boyouncy force and gravitational force , as we are not just extracting mass we are not changing its volume so our Fb remains same but Fg changes
And for that i have solved it number of times
At the end i reached at some value of radius of mass to be extracted from that solid ball
After extraction of mass i have now hollow shell, now i have to fill it with other materials having density of 630g/m3, and i have inner radius known , with the help of that i found mass of that added material,
Then i found cumulative mass of solid ball (having two materials, inner material of density 630g/m3, coated with other materials of density 2700g/m3)
After finding commutative mass i found Fg with the help of it and then checked for
Fb = Fg
But with this method i came close, mean i got idea how should i have to solve this problem then i generated a excel sheet applied all the formulas i used and i varried the radius with the help of excel sheet i generated i found that at these radius 0.025m and 0.024m my result are same for Fb=Fg
This is how i did it

 

[haruspex]

Ok what first i did is i have assumed that i have a solid ball having some radius, with the help of that radius i calculated volume , then with density and volume i found mass of that ball

Then i used
Fb = Fg

Till this point the values of these two doesn't comes equal to each other, so now i have to etract some mass from that solid ball and mass should be equal to that with the help of which we attain equality between boyouncy force and gravitational force , as we are not just extracting mass we are not changing its volume so our Fb remains same but Fg changes
And for that i have solved it number of times
At the end i reached at some value of radius of mass to be extracted from that solid ball
After extraction of mass i have now hollow shell, now i have to fill it with other materials having density of 630g/m3, and i have inner radius known , with the help of that i found mass of that added material,
Then i found cumulative mass of solid ball (having two materials, inner material of density 630g/m3, coated with other materials of density 2700g/m3)
After finding commutative mass i found Fg with the help of it and then checked for
Fb = Fg
But with this method i came close, mean i got idea how should i have to solve this problem then i generated a excel sheet applied all the formulas i used and i varried the radius with the help of excel sheet i generated i found that at these radius 0.025m and 0.024m my result are same for Fb=Fg
This is how i did it

That's not a very direct method. Try the method I outlined at the end of post #7.

 

[Aqib409]

That's not a very direct method. Try the method I outlined at the end of post #7.

I tried it, but didnt found any expression or formula to go with it

 

[haruspex]

I tried it, but didnt found any expression or formula to go with it

In terms of the variables I defined, what are the volumes of the particle and its coating? What is the combined mass?

 

[Aqib409]

 

[haruspex]

View attachment 240579

Right.
It's a bit confusing the way you are using r₁ for the inner radius but ρ₁ for the coating density, but no matter.
There is some cancellation you can do, and regroup into r₁ terms and r₂ terms since these are your unknowns.